Metamath Proof Explorer
Description: The order of a group element is an integer. (Contributed by Mario
Carneiro, 28-Feb-2014)
|
|
Ref |
Expression |
|
Assertion |
odzcl |
⊢ ( ( 𝑁 ∈ ℕ ∧ 𝐴 ∈ ℤ ∧ ( 𝐴 gcd 𝑁 ) = 1 ) → ( ( odℤ ‘ 𝑁 ) ‘ 𝐴 ) ∈ ℕ ) |
Proof
Step |
Hyp |
Ref |
Expression |
1 |
|
odzcllem |
⊢ ( ( 𝑁 ∈ ℕ ∧ 𝐴 ∈ ℤ ∧ ( 𝐴 gcd 𝑁 ) = 1 ) → ( ( ( odℤ ‘ 𝑁 ) ‘ 𝐴 ) ∈ ℕ ∧ 𝑁 ∥ ( ( 𝐴 ↑ ( ( odℤ ‘ 𝑁 ) ‘ 𝐴 ) ) − 1 ) ) ) |
2 |
1
|
simpld |
⊢ ( ( 𝑁 ∈ ℕ ∧ 𝐴 ∈ ℤ ∧ ( 𝐴 gcd 𝑁 ) = 1 ) → ( ( odℤ ‘ 𝑁 ) ‘ 𝐴 ) ∈ ℕ ) |